Proof. Assume towards contradiction that $G$ is a simple disconnected graph on $12$ vertices with $56$ edges. This means the graph is a union of components, i.e. $$G=g_1\cup g_2\cup \cdots \cup g_k$$ Where $2\le k\le 12$.

Letting $v_k=|V(g_k)|$, we get that $$|E(g_k)|\le\sum_{i=1}^{v_k-1} i=\frac{v_k(v_k-1)}{2}$$ Therefore $$\begin{align*} |E(G)|&\le\sum_{i=1}^{k}\frac{v_i(v_i-1)}{2} \\ &=\frac{1}{2}\sum_{i=1}^{k}\left(v_i^2-v_i\right) \\ &=\frac{1}{2}\sum_{i=1}^{k}v_i^2-\frac{1}{2}\sum_{i=1}^{k}v_i \\ \end{align*}$$ The sum $\sum_{i=1}^{k}v_i$ is the sum of all of the vertices, which is $12$. $$\begin{align*} &=\frac{1}{2}\sum_{i=1}^{k}v_i^2-\frac{1}{2}(12) \\ &=\frac{1}{2}\sum_{i=1}^{k}v_i^2-6 \\ \end{align*}$$

Letting $v_i\le v_j$ we get that $$\begin{align*} v_i^2+v_j^2&<v_i^2+v_j^2+2v_j-2v_i+2 \\ &=v_i^2-2v_i+1+v_j^2+2v_j+1 \\ &=(v_i-1)^2+(v_j+1)^2 \end{align*}$$ This shows that in order to maximize the sum we should move the vertices into the larger components, even to the point of deleting smaller components. This leads us to the following. $$\begin{align*} |E(G)|&\le\frac{1}{2}\sum_{i=1}^{2}v_i^2-6 \\ &=\frac{1}{2}[v_1^2+v_2^2]-6 \\ &=\frac{1}{2}[v_1^2+(12-v_1)^2]-6 \\ \end{align*}$$ Using previous logic this is maximized for $v_1=1$. $$\begin{align*} |E(G)|&\le\frac{1}{2}[1^2+(12-1)^2]-6 \\ &=\frac{1}{2}[122]-6 \\ &=55 \\ \end{align*}$$ In conclusion, $|E(G)|\le55$, i.e. $G$ can have at most $55$ edges, which is a contradiction.